Question

In a shipping company distribution center, an open cart of mass 50.0 kg is rolling
to the left at a speed of 5.00 m/s. Ignore friction between the cart and the floor. A 15.0-kg
package slides down a chute that is inclined at 37o from the horizontal and leaves the
end of the chute with a speed of 3.00 m/s. The package lands in the cart and they roll
together. If the lower end of the chute is a vertical distance of 4.00 m above the bottom
of the cart, what are (a) the speed of the package just before it lands in the cart and (b)
the final speed of the cart?

I have a question for part (a).

I first set up (a) as >>> (K2 +U2) = (K1 +U1)
>> (.5)(m)(v^2) + mgh = (.5)(m)(v^2) + mgh

>> (.5)(15)(3^2) + (15)(9.8)(4) = (.5)(15)(VelocityB^2) + (15)(9.8)(0)

>> the .5 and 15 from both sides cross out so

>> (3^2) + (15)(9.8)(4) = VelocityB^2

>> in my book though where i have 15 just above they put 2. Why is this? I thought the mass of the object is 15 kg so 15 goes there not 2.

Answer 1